LT1375/ LT1376
U
W
U U
APPLICATIONS INFORMATION
INDUCTOR VALUE
Output current where continuous mode is needed:
Unlike buck converters, positive-to-negative converters
cannot use large inductor values to reduce output ripple
voltage. At 500kHz, values larger than 25µH make almost
no change in output ripple. The graph in Figure 19 shows
peak-to-peak output ripple voltage for a 5V to –5V con-
verter versus inductor value. The criteria for choosing the
inductor is therefore typically based on ensuring that peak
switch current rating is not exceeded. This gives the
lowest value of inductance that can be used, but in some
cases (lower output load currents) it may give a value that
creates unnecessarily high output ripple voltage. A com-
promise value is often chosen that reduces output ripple.
As you can see from the graph, large inductors will not
give arbitrarily low ripple, but small inductors can give
high ripple.
2
V
2 I
( ) ( P)
IN
ICONT
=
4 V + V
V + V + V
IN OUT F
(
OUT)(
)
IN
Minimum inductor discontinuous mode:
2 V
I
(
OUT)( OUT
)
LMIN
=
2
f I
( )( P)
Minimum inductor continuous mode:
V V
( )(
)
IN OUT
LMIN
=
V
+ V
150
F
(
)
OUT
5V TO –5V CONVERTER
OUTPUT CAPACITOR
2 f V + V
I −I
1+
( )(
)
IN
OUT
P
OUT
V
IN
ESR = 0.1Ω
120
90
60
30
0
For the example above, with maximum load current of
0.25A:
I
= 0.25A
LOAD
I
= 0.1A
LOAD
5 2 1.5 2
( ) (
)
ICONT
=
= 0.37A
4 5 + 5 5 +5 +0.5
(
)(
)
0
5
10
15
20
25
INDUCTOR SIZE (µH)
This says that discontinuous mode can be used and the
minimum inductor needed is found from:
1375/76 F19
Figure 19. Ripple Voltage on Positive-to-Negative Converter
2 5 0.25
( )(
)
LMIN
=
= 2.2µH
2
The difficulty in calculating the minimum inductor size
needed is that you must first know whether the switcher
will be in continuous or discontinuous mode at the critical
point where switch current is 1.5A. The first step is to use
the following formula to calculate the load current where
the switcher must use continuous mode. If your load
current is less than this, use the discontinuous mode
formula to calculate minimum inductor needed. If load
current is higher, use the continuous mode formula.
500 •103 1.5
(
)
In practice, the inductor should be increased by about
30% over the calculated minimum to handle losses and
variations in value. This suggests a minimum inductor of
3µH for this application, but looking at the ripple voltage
chartshows thatoutputripplevoltagecouldbereducedby
a factor of two by using a 15µH inductor. There is no rule
25
Linear [ Linear ]
